ADVANCED SUBSIDIARY (AS) General Certificate of Education Mathematics Assessment Unit F1. assessing. Module FP1: Further Pure Mathematics 1

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1 ADVACED SUBSIDIARY (AS) Geneal Cetificate of Education 15 Mathematics Assessment Unit F1 assessing Module F1: Futhe ue Mathematics 1 [AMF11] WEDESDAY 4 UE, MRIG MAR SCHEME F

2 GCE ADVACED/ADVACED SUBSIDIARY (AS) MATHEMATICS Intoduction The mak scheme nomally povides the most popula solution to each question. the solutions given by candidates ae evaluated and cedit given as appopiate; these altenative methods ae not usually illustated in the published mak scheme. The maks awaded fo each question ae shown in the ighthand column and they ae pefixed by the lettes M, W and MW as appopiate. The key to the mak scheme is given below: M W indicates maks fo coect method. indicates maks fo woking. MW indicates maks fo combined method and woking. The solution to a question gains maks fo coect method and maks fo an accuate woking based on this method. Whee the method is not coect no maks can be given. A late pat of a question may equie a candidate to use an answe obtained fom an ealie pat of the same question. A candidate who gets the wong answe to the ealie pat and goes on to the late pat is natually unawae that the wong data is being used and is actually undetaking the solution of a paallel poblem fom the point at which the eo occued. If such a candidate continues to apply coect method, then the candidate s individual woking must be followed though fom the eo. If no futhe eos ae made, then the candidate is penalised only fo the initial eo. Solutions containing two o moe woking o tansciption eos ae teated in the same way. This pocess is usually efeed to as followthough making and allows a candidate to gain cedit fo that pat of a solution which follows a woking o tansciption eo. ositive making: It is ou intention to ewad candidates fo any demonstation of elevant knowledge, skills o undestanding. Fo this eason we adopt a policy of following though thei answes, that is, having penalised a candidate fo an eo, we mak the succeeding pats of the question using the candidate s value o answes and awad maks accodingly. Some common examples of this occu in the following cases: (a) a numeical eo in one enty in a table of values might lead to seveal answes being incoect, but these might not be essentially sepaate eos; (b) eadings taken fom candidates inaccuate gaphs may not agee with the answes expected but might be consistent with the gaphs dawn. When the candidate miseads a question in such a way as to make the question easie only a popotion of the maks will be available (based on the pofessional judgement of the examining team) F

3 AVAIABE MARS F AVAIABE MARS F 1 (i) (ii) Since, then R is an eigenvecto of R with a coesponding eigenvalue of MW (iii) R # 4 R RR 6 4 AVAIABE MARS 8

4 (i) x + y 4x8y + 1 & ^xh + ^y 4h 1 AVAIABE MARS Hence the cicle has cente (, 4) 5 4 & Gadient of adius 1 1 Altenative solution: dy dy x+ y 4 8 d x dx dy 4 x ` At pt (1,5) dx ( y 8) d y 4( 1) 6 dx 1 8 Theefoe the gadient of the tangent is & Equation of tangent is y x + c Using the point ( 1, 5) gives 5 + c Hence c 8 and the equation of the tangent is y x + 8 (ii) y ( 1, ( 1,5) 5) C (, 4) C (, 4) Q x Fom the diagam, to C is ight and 1 down By symmety, then C to Q is also ight and 1 down Hence Q is the point (5, ) Since y x + c, then using (5, ) gives 15 + c & c 1 Hence the othe tangent is y x F 4

5 (ii) Altenative Solution The paallel tangent is y x + c Substitute to give x + ( x + c) 4x8 ^x + ch + 1 & x + 9x + 6cx + c 4x4x8c+ 1 & 1 x + x(6c8) + ( c 8c+ 1) Fo a tangent, then B 4AC Hence, (6c8) 4( c 8c+ 1) AVAIABE MARS & 6c 6c c c+ 4 & 4c + 16c 84 & c + 4c 96 & ( c+ 1) ( c 8) & c 1, 8 Hence the othe tangent is y x F 5

6 (i) a fo nonunique solution. 1 a Hence a(4a 1) 1 ( a + ) (1 + 8) & 4a aa 18 & 4a a & a a1 & ( a+ )(a 5) & a, a 1 AVAIABE MARS (ii) a gives a nonunique solution Setting up equations x + y z 6 x + 4y + z x + y z 5 Since and ae contadictoy then thee is no solution. (iii) Setting up equations x + y z 5 x+ 4y + z x+ y z 5 Since and ae identical, then thee ae infinitely many solutions. x + y z 5 x + 8y + z 4 Adding gives 9y 9 Hence y 1 & x + 4+ z & x + z Theefoe, the geneal solution is (t, 1, t) F 6

7 4 Closue s p q p + sq q + sp d nd n d n s q p sp + q sq + p w v d n, whee w p+ sqand v w v q+ sp It is also necessay to show that w v w p + psq + s q v q + psq + s p and hence ae not equal, since p q, s Hence the poduct is of the same fomat and closue exists. AVAIABE MARS Identity Identity fo multiplication is 1 d n 1 Test 1 s s d nd n d n 1 s s and s 1 s d nd n d n s 1 s Since 1 d n is of the same fomat as 1 d s s n, then an identity exists. Altenative Solution a b s s d nd n d n b a s s Hence a + sb and sa + b s & a+ sb and sa+ sb s Subtacting gives ^ s h a s & ^ s h^a1h Hence a 1 since! s Theefoe, b Hence identity is d 1 n F 7

8 Invese s The invese d n unde matix multiplication is s 1 s e o s s AVAIABE MARS Since s, then the invese exists and the matix is of the same s fomat as d n s Since the associative law is also tue, then S foms a goup unde matix multiplication F 8

9 cos i sin i 5 (a) The equied matix is d n sin i cos i 1 1 i 45c & matix is f 1 1 p AVAIABE MARS (b) Unde the mapping, the line y mx is mapped to y mx 1 t x & d nd n d n 6 mt mx & t + mt x 6t + mt mx Dividing gives + m 1 6+ m m Expanding gives m m 6 + m & m + 5m 6 & ( m+ 6) ( m 1) & m 6, m 1 W F 9

10 6 (a) (i) (a + bi) 5 + 1i & a + abi b 5 + 1i Equating eal tems: a b 5 Equating imaginay tems: ab 1 AVAIABE MARS Rewite 6 to give b a Substitute into to give 6 a ` a j 5 4 & a + 5a 6 & ^a + 9h ^a 4h & a! & b! (ii) Use quadatic fomula to obtain ^4ih! ^4ih 4^55ih z ^4ih! 168i1+ i & z ^4ih! 5+ 1i & z ^4 ih! ^+ ih & z 6 i 4i & z +, Hence z + i, 1 i F 1

11 (b) (i) Cicle, cente (, ) and adius MW AVAIABE MARS y Q C (, C (, ) ) θ θ S x (ii) The minimum/maximum values of ag w occu at the points /Q, whee /Q ae tangents to the cicle. C + C 18 1 θ & sin i C 18 & i 6 tan a 1 & a S 4 Minimum value of ag w is Maximum value of ag w is ` 6 j Hence 1 5 Gag w G 1 a θ C Total F 11

0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, maximum raw mark 80

0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, maximum raw mark 80 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Intenational Geneal Cetificate of Seconday Education MARK SCHEME fo the Octobe/Novembe 009 question pape fo the guidance of teaches 0606 ADDITIONAL MATHEMATICS